Updated By SaraNextGen

On March 11, 2024, 11:35 AM

Question 7.62:

A 0.02 M solution of pyridinium hydrochloride has pH = 3.44. Calculate the ionization constant of pyridine

Answer:

pH = 3.44

We know that,

pH = – log [H⁺]

Question 7.63:

Predict if the solutions of the following salts are neutral, acidic or basic:

NaCl, KBr, NaCN, NH₄NO₃, NaNO₂ and KF

Answer:

(i) NaCl:

Therefore, it is a neutral solution.

(ii) KBr:

Therefore, it is a neutral solution.

(iii) NaCN:

Therefore, it is a basic solution.

(iv) NH4NO3

Therefore, it is an acidic solution.

(v) NaNO2

Therefore, it is a basic solution.

(vi) KF

Therefore, it is a basic solution.

Question 7.64:

The ionization constant of chloroacetic acid is 1.35 × 10^–3. What will be the pH of 0.1M acid and its 0.1M sodium salt solution?

Answer:

It is given that Ka for ClCH₂COOH is 1.35 × 10^–3.

ClCH₂COONa is the salt of a weak acid i.e., ClCH₂COOH and a strong base i.e., NaOH.

Question 7.65:

Ionic product of water at 310 K is 2.7 × 10^–14. What is the pH of neutral water at this temperature?

Answer:

Ionic product,

Hence, the pH of neutral water is 6.78.

Question 7.66:

Calculate the pH of the resultant mixtures:

a) 10 mL of 0.2M Ca(OH)₂ + 25 mL of 0.1M HCl

b) 10 mL of 0.01M H₂SO₄ + 10 mL of 0.01M Ca(OH)₂

c) 10 mL of 0.1M H₂SO₄ + 10 mL of 0.1M KOH

Answer:

(a) 

Thus, excess of   = .0015 mol

(b) 

Since there is neither an excess of  or , the solution is neutral. Hence, pH = 7.

(c) 

Excess of  = .001 mol

= 1.30

Question 7.67:

Determine the solubilities of silver chromate, barium chromate, ferric hydroxide, lead chloride and mercurous iodide at 298K from their solubility product constants given in Table 7.9 (page 221). Determine also the molarities of individual ions.

Answer:

(1) Silver chromate:

Let the solubility of  be s.

Molarity of  = 2s = 2 × 0.65 × 10^–4 = 1.30 × 10^–4 M

Molarity of  = s = 0.65 × 10^–4 M

(2) Barium chromate:

Let s be the solubility of 

Thus,   = s and   = s

Molarity of  = Molarity of 

(3) Ferric hydroxide:

Let s be the solubility of 

Molarity of 

Molarity of 

(4) Lead chloride:

Let KSP be the solubility of 

Molarity of 

Molarity of chloride =

(5) Mercurous iodide:

Let s be the solubility of 

Molarity of 

Molarity of 

Question 7.68:

The solubility product constant of Ag₂CrO₄ and AgBr are 1.1 × 10^–12 and 5.0 × 10^–13 respectively. Calculate the ratio of the molarities of their saturated solutions.

Answer:

Let s be the solubility of Ag₂CrO_4.

Let s´ be the solubility of AgBr.

Therefore, the ratio of the molarities of their saturated solution is 

Question 7.69:

Equal volumes of 0.002 M solutions of sodium iodate and cupric chlorate are mixed together. Will it lead to precipitation of copper iodate? (For cupric iodate K_sp = 7.4 × 10^–8).

Answer:

When equal volumes of sodium iodate and cupric chlorate solutions are mixed together, then the molar concentrations of both solutions are reduced to half i.e., 0.001 M.

Then,

Now, the solubility equilibrium for copper iodate can be written as:

Ionic product of copper iodate:

Since the ionic product (1 × 10^–9) is less than Ksp (7.4 × 10^–8), precipitation will not occur.

Question 7.70:

The ionization constant of benzoic acid is 6.46 × 10^–5 and K_sp for silver benzoate is 2.5 × 10^–13. How many times is silver benzoate more soluble in a buffer of pH 3.19 compared to its solubility in pure water?

Answer:

Since pH = 3.19,

Let the solubility of C₆H₅COOAg be x mol/L.

Then,

Thus, the solubility of silver benzoate in a pH 3.19 solution is 1.66 × 10^–6 mol/L.

Now, let the solubility of C₆H₅COOAg be x’ mol/L.

Hence, C₆H₅COOAg is approximately 3.317 times more soluble in a low pH solution.

Question 7.71:

What is the maximum concentration of equimolar solutions of ferrous sulphate and sodium sulphide so that when mixed in equal volumes, there is no precipitation of iron sulphide? (For iron sulphide, K_sp = 6.3 × 10^–18).

Answer:

Let the maximum concentration of each solution be x mol/L. After mixing, the volume of the concentrations of each solution will be reduced to half i.e., .

If the concentrations of both solutions are equal to or less than 5.02 × 10^–9 M, then there will be no precipitation of iron sulphide.

Question 7.72:

What is the minimum volume of water required to dissolve 1g of calcium sulphate at 298 K? (For calcium sulphate, K_sp is 9.1 × 10^–6).

Answer:

Let the solubility of CaSO₄ be s.

Molecular mass of CaSO₄ = 136 g/mol

Solubility of   in gram/L = 3.02 × 10^–3 × 136

= 0.41 g/L

This means that we need 1L of water to dissolve 0.41g of CaSO₄

Therefore, to dissolve 1g of CaSO₄ we require  of water.

Question 7.73:

The concentration of sulphide ion in 0.1M HCl solution saturated with hydrogen sulphide is 1.0 × 10^–19 M. If 10 mL of this is added to 5 mL of 0.04 M solution of the following: FeSO₄, MnCl₂, ZnCl₂ and CdCl₂. in which of these solutions precipitation will take place?

Answer:

For precipitation to take place, it is required that the calculated ionic product exceeds the Ksp value.

Before mixing:

After mixing:

This ionic product exceeds the Ksp of Zns and CdS. Therefore, precipitation will occur in CdCl₂ and ZnCl₂ solutions.